molecule A would form a 6 electron aromatic system upon proton abstraction, so it is basic according to the 4N+2 rule. Molecule B would form an antiaromatic system with 10pi electrons. molecule C would provide a 14 pi electron system, which is also aromatic as you said...molecule D would again form an antiaromatic system with 10 pi electrons. in comparing A and C i would say C is more acidic because that carbanion can be resonated all throughout the things, while molecule A only provides a couple positions

Forget that you know the pKa's for a minute and consider this situation: you have a solution of the conjugate base of fluorene (Fl-) and treat it with cyclopentadiene (Cp-H)

Fl- + Cp-H Fl-H + Cp-arom. non-arom arom. arom.

It's common practice to compare the relative stabilities of the charged species of an acid-base equilibrium, but that doesn't give the whole picture. Here, you also have to compare the stability of Cp-H to Fl-H

If I had been asked by a professor on the spot, I would have said Fluorene. It has more resonance forms, and should make the Fl-H bond weaker, so more acidic. However, what about considering any possible electron-donating character of the benzene rings? Enough to destabilize the anion of the Fl- to increase the pKa enough?

Right. If you deprotonate fluorene, then the negative charge can resonante throughout the whole molecule. Much more delocalization than in cyclopentadiene. This will help stabilize the negative charge, so you would think that fluroene should be more acidic.

I am wondering if the inductive effects of the benzene rings are enough to cause the decrease in acidity, or what else may be causing that.